Delays
Hitay Özbay1and A. Nazli Günde¸s2 1
Dept. Electrical & Electronics Eng., Bilkent Univ., Ankara, 06800 Turkey, hitay@bilkent.edu.tr
2
Dept. Electrical & Computer Eng., Univ. of California, Davis, CA 95616, U.S.A., angundes@ucdavis.edu
Summary. Consider a stabilizing controller C1for a given plant P . If C1and P do not have any zeros at the origin, then one can use a cascade connected PI (proportional plus integral) controller Cpiwith C1 and keep the feedback system stable. In this work we examine the
allowable range of the integral action gain in Cpi, and discuss how C1should be chosen to
maximize this range for systems with time delays.
1
Introduction
In the design of feedback controllers it is often desirable to use an integrator to be able to track constant reference signals. For example, internal model principle says that the controller must include a copy of the reference signal (or disturbance) generator in order to have a robust tracking (or disturbance rejection), see e.g. [2, 4, 7]. Typically, the reference generator Gr(s)
is an unstable system: an integrator (resp. oscillator) if the reference is a constant (resp. a sinusoidal signal). One way to achieve robust asymptotic tracking (or disturbance rejection) is to append Grto the plant P and then design a controller Cofor the combined “plant” GrP.
Thus C = CoGris a stabilizing controller for P and it achieves the performance objectives,
see e.g. [1, 16] for more details.
In this chapter we consider the dual problem: first design a stabilizing controller for the plant, then append a PI term to this controller. A similar problem has been discussed in [3] for finite dimensional systems. Briefly, the problem we deal with can be stated as follows: let
C1be a stabilizing controller for a time delay system P , and append (in the form of a cascade connection) Cpi(s) = (s+ksi) to C1. Note that the proportional gain of the PI controller is
set to unity; this is without loss of generality since a non-unity gain can be absorbed into C1. Assume that P and C1do not have any zeros at the origin. Then, there exists kisuch that the
feedback system is stable. We examine the range of allowable ki, and discuss the problem of
designing an optimal C1so that this range is maximized.
We should indicate that rather than the cascade PI-controller connection to be discussed here, a two-stage parallel connection of controllers is also possible. For example, as before, let C1be a a stabilizing feedback controller for a given plant P . If the PI part of the controller,
This work was supported in part by TÜB˙ITAK (grant no. EEEAG-105E156).
J.J. Loiseau et al. (Eds.): Topics in Time Delay Systems, LNCIS 388, pp. 197–207. springerlink.com Springer-Verlag Berlin Heidelberg 2009c
Cpi, is a stabilizing controller for the new plant P (I + C1P )−1, then the parallel connection
of the controller, Cpi+ C1, is a stabilizing controller for the original plant P , see [9, 14,
16]. One can study the problem choosing the best C1 so that the allowable range of ki is
maximum. We leave this problem aside, because the techniques to be used in such a study would be similar to the approach taken in this chapter for the cascade connection of the controllers.
This chapter is organized as follows: stability of the feedback system under cascade con-nection of the PI controller is investigated in Section 2. Design of C1, maximizing the allow-able range of the integral gain, is discussed in Section 3. Concluding remarks are made in Section 4.
Notation used here is standard. In particular, the norm sign · stands for the H∞norm
· ∞whenever the argument is inH∞.
2
Feedback System Stability Under Cascade Connection of the
PI Controller
Consider the feedback system shown in Figure 1 with an r input r output plant whose r× r transfer matrix is P (s). The r× r controller transfer matrix is C(s). Assume that P is full rank. The feedback system is said to be stable if C(1 + P C)−1, P C(1 + CP )−1, C(I + P C)−1P, P (1 + CP )−1are inHr×r
∞ . In this case, we say C∈ S(P ), where S(P ) is the
set of all controllers stabilizing the feedback system with plant P .
Fig. 1. Feedback System
Let C1be inS(P ) and consider the cascade connection C = C1Cx for some Cx. The
result stated below as Theorem 1 addresses the following question: Is the closed-loop system still stable if C = C1Cx, i.e. do we have C ∈ S(P )?
Theorem 1. Let P be a given r× r plant and let C1∈ S(P ). Assume that P and C1are full
rank and define the complementary sensitivity function for the feedback system with C = C1
as T1:= P C1(I + P C1)−1. Then, we have the following two results:
a) If Cy:= Cx− I stabilizes T1∈ Hr∞×r, then
C = C1Cx∈ S(P ) . (1) b) Let P and C1 have no transmission-zeros at the origin. Choose any ˆKP, ˆKD ∈ Rr×r,
Ψ (s) =T1(s)T1(0) −1− I s + T1(s) ( ˆKP+ ˆ KDs τ s + 1). (2) Then for ρ∈ R+ satisfying
ρ < Ψ −1=: ψ−1 (3)
the controller C = C1Cpid ∈ S(P ) , where Cpid is a PID-controller given by Cpid =
I + ˆCpid, where ˆ Cpid= ρ ˆ KP + T1(0)−1 s + ˆ KDs τ s + 1 . (4)
For ˆKD= 0, (4) becomes a PI-controller.
Proof of Theorem 1 is given in the Appendix. By part (a) of this theorem the stabilizing controller C = C1Cx gives rise to the following complementary sensitivity Tx = P C(I +
P C)−1:
Tx = P C1(I + CxP C1)−1Cx= P C1(I + P C1+ CyP C1)−1Cx
= T1(I + CyT1)−1(I + Cy) = (I + T1(I− Cx))−1T1Cx. (5)
If Cx= Cpidas in (4) of Theorem 1-(b), then Txin (5) becomes
T = (I + T1Cˆpid)−1T1Cpid, (6)
which can be expressed as
T = ( s s + ρI + T1(s) ˆCpid(s) s s + ρ) −1T1(s)( s s + ρI + s s + ρCˆpid) = (I + ρs s + ρΨ (s) ) −1T1(s) ρs s + ρ(I + ρ s( ˆKP+ T1(0)−1 s + ˆ KDs τ s + 1)). (7) Therefore, T (0) = I and T ≤ (I + ρs s + ρΨ ) −1 · T1(s) s s + ρ(I + ˆCpid) . Writing (I + s+ρρs Ψ )−1= I− (I +s+ρρs Ψ )−1 ρss+ρΨ, we obtain (I + ρs s + ρΨ (s) ) −1 ≤ 1 + ρ ψ (I + ρs s + ρΨ ) −1
and hence, (I +s+ρρs Ψ (s) )−1 ≤ (1 − ρ ψ)−1, and
T ≤ 1
1− ρ ψ T1(s)
s
s + ρ(I + ˆCpid(s) ). (8)
Now suppose that in the PID-controller ˆCpid we choose ˆKP = 0, ˆKD= 0. Then by (4), the
PI-controller is
Cpi(s) = I +
ρ T1(0)−1
s ,
where ρ∈ R+ satisfies (3), i.e.,
ρ <T1(s)T1(0)
−1− I
s
−1=: ψ−1
In this case, the upper-bound on T given in (8) becomes T ≤ 1 1− ρ ψo T 1T1(0)−1 · sT1(0) + ρI s + ρ .
In particular, if T1(0) = I, i.e. C1and/or P contain a pole at s = 0, then
T ≤ 1
1− ρ ψo T
1.
>From the above discussion we see that if ρ ψo! 1 then the upper bound of T is close to
T1.
3
Design of
C
1Maximizing the Integral Action Gain
In this section we discuss the design of C1for the largest allowable range of ρ, (9), for a class of single input single output (SISO) plants with time delays. In this case, from (9) we see that
C1should be designed to minimize
ψ =T1(s)T
−1
1 (0)− 1
s ∞, (10)
where T1 = P C1(1 + P C1)−1and C1∈ S(P ).
Solution of this problem will be obtained below in two steps: (i) first we solve the problem for stable plants, then (ii) we extend this solution to cover unstable plants case. In both steps we begin with inner-outer and coprime factorizations of given P , then we solve anH∞ opti-mization problem. Inner-outer factorizations require findingC+roots of a quasi-polynomial, for which several algorithms exist by now, see e.g. [5, 12, 17] and their references. Using these algorithms and the methods developed for theH∞control of general infinite dimen-sional systems, (see e.g. [6] and [8]) we can solve the problem in step (i) for a large class of time delay systems. We will see that the extension (ii) to unstable plants, with finitely many poles inC+, involves a parameterization of all suboptimal solutions of the problem in (i), and the use of Nevanlinna-Pick interpolation. The mathematical tools for these problems can be found in [6, 11, 15, 18].
3.1 Stable plants
In this section we consider stable SISO plants whose inner-outer factorizations are in the form
P = PiPowhere Piis inner (all-pass) with Pi(0) = 1, and Pois outer (minimum-phase).
Example.Consider the plant with input/output delay, h > 0, and internal delays
P (s) = e−hs s + 2 (s + 1) + 2(s− 1)e−2s (s + 3) + e−3s . (11)
Then, the following is an inner-outer factorization:
Pi(s) =−e−hs (s + 1) + 2(s− 1)e−2s 2(s + 1) + (s− 1)e−2s Po(s) = −1 s + 2 2(s + 1) + (s− 1)e−2s (s + 3) + e−3s
Clearly, Pi(0) = 1, the poles and zeros of Piare symmetric around the Im-axis, and Po
contains no poles or zeros in the right half plane. The set of all stabilizing controllers is parameterized as
S(P ) = {Q/(1 − P Q) : Q ∈ H∞and P Q= 1}.
Therefore, C1must be in the form C1= Q1/(1− P Q1), where Q1∈ H∞is free. Let
Q1(s) = Po−1(s)
Qi(s)
(1 + εs) (12)
where Qi ∈ H∞is the free parameter, ε > 0, and " is the relative degree of Po. Then we
have
T1(s) = P (s)Q1(s) = Pi(s)
Qi(s)
(1 + εs).
Hence the problem of maximizing the allowable range of ρ reduces to finding
ψo = inf Qi∈H∞( Pi(s) (1 + εs)Qi(s)Q −1 i (0)− 1)/s∞ =( Pi(s) (1 + εs)Qi,opt(s)Q −1 i,opt(0)− 1)/s∞ (13)
and the corresponding optimal Qi,opt∈ H∞solving this problem. Note that optimal solution is not unique: if Qi,ois a solution of (13), then so is KQi,o, for any non-zero constant K.
Therefore, we define the normalized free parameter Qi(s) = Qi(s)Q−1i (0), and try to find
optimal Qiin the problem (14) defined below. First letHo∞={ Qi∈ H∞ : Qi(0) = 1}.
Note that ψo = inf Qi∈Ho ∞ ( Pi(s) (1 + εs)Qi(s)− 1)/s∞ ≥ inf Qi∈H∞ ( Pi(s) (1 + εs)Qi(s)− 1)/s∞=: ψo.
But the optimal solution of the problem defining ψomust lie inHo∞, because ((1+εs)Pi(s)Qi(s)−
1)/sis inH∞only if Qi(0) = 1. Therefore, ψo= ψoand Qi,optis the optimal solution of
ψo= inf
Qi∈H∞
( Pi(s)
(1 + εs)Qi(s)− 1)/s∞. (14)
The problem (14) is a one-blockH∞optimization problem, which can be seen as equivalent to a weighted sensitivity minimization for a stable plant with the sensitivity weigh being an integrator. For a general inner function Pi, the problem (14) can be solved using the
tech-niques developed for theH∞control of infinite dimensional systems, see e.g. [6, 10, 13] and their references. It turns out that the optimal Qiis in the form
Qi,opt(s) = (1 + εs) (1 + δs)(+1) 1 + ψ2os2 Pi(s) + ψos (15) where δ→ 0 and ψois the largest value of ψ > 0 for which we have
Choosing Q1 as in (12) with Qi = K1Qi,opt, for an arbitrary K1 = 0, and defining the controller C1= Q1/(1− P Q1), we obtain C1(s) = K1Po−1(s) (1+δs)+1(ψos+Pi(s)) 1+ψ2os2 − K1Pi(s) . (17)
ρ = kiK1 < ψ−1o . Hence, depending on the gain K1used in C1we get an allowable range
for ki,
|ki| < ψo−1/|K1| .
Note that ψois invariant and completely determined by the inner part Pi(s)of the plant.
Another interesting problem in this context is to investigate PD (proportional plus deriva-tive) type of Qi(s) = (1 + kds)in (14). More precisely,
ψpd:= inf kd∈R( Pi(s) (1 + εs)(1 + kds)− 1)/s∞= infk d∈R f (s)− 1 s + kdf (s)∞. (18) where f (s) = Pi(s)
(1+εs). The function f is inH∞and f (0) = 1. This problem has been
studied in the context of resilient PD controller design in [14] and a closed form expression is obtained for the optimal kd.
Example (Revisited) For the example given in (11), the equation (16) can be written as
e−jh/ψm(j/ψ) 1− m(−j/ψ)/2 1− m(j/ψ)/2 =−j
where m(s) = e−2s (1−s)(1+s). Since m(s) is inner we have m(j/ψ) = e−jθm, where θm =
2/ψ + 2 tan−1(1/ψ). We also have1− m(−j/ψ)/2 1− m(j/ψ)/2 = e −jθ, where θ = 2 tan−1 sin(θm) 2−cos(θm)
. Therefore ψ−1o is the smallest x satisfying
(1 +h
2)x + tan
−1(x) + tan−1Ω(x) = π
4 (19)
where Ω(x) = sin(2(x + tan
−1(x)))
2− cos(2(x + tan−1(x))).
We should also note that if we change the inner part of the plant to an input/output de-lay, Pi(s) = e−hs (i.e. consider m(s) = 1), then from (16) we get ψo−1 = π/2h, which
is precisely the gain margin of the feedback system whose open loop transfer function is
e−hs/s.
When we consider a PD type of Qi(s) = (1 + kds), the solution of (18) gives optimal
ψpdoptand the corresponding kdopt. Figure 2 illustrates how the optimal ψo, ψpdoptand kdoptvary
with h. Note from this figure that the use of PD term does not lead to significant performance degradation (reduction in the largest allowable kirange) compared to the use of optimal Qi
of (15).
3.2 Extension to Unstable Plants
Now consider unstable SISO plants factorized as
P (s) = N (s) Di(s)
10−2 10−1 100 101 102 10−2 10−1 100 h h versus ψo −1 , ψpd −1 , and 1/kdopt ψ o −1 1/k d opt ψpd−1
Fig. 2. h versus ψ−1o , ψ−1pd and 1/k
opt
d .
where No(s)is outer, Ni(s)and Di(s)are inner with Dibeing a finite Blaschke product
(i.e. the plant has finitely many poles inC+, and it has no poles on the Im-axis). As before, we will assume that Ni(0) = 1. For this type of plants C1∈ S(P ) if and only if
C1(s) =X(s) + Di(s)Q1(s) Y (s)− N(s)Q1(s) for some Q1∈ H∞ (21) where X, Y ∈ H∞satisfy Y (s) =1− N(s)X(s) Di(s) . (22)
Let p1, . . . , pnbe the zeros of Di(s), i.e. poles of P (s) inC+, and for simplicity of the
exposition assume that they are distinct. Then, Y ∈ H∞if and only if the function X∈ H∞
satisfy X(pi) = 1/N (pi), i = 1, . . . , n. If we use C1 in the form of (21) as the initial
stabilizing controller for the plant P , then
T1(s) = N (s)(X(s) + Di(s)Q1(s)). Therefore ψ =(T1(s)T1−1(0)− 1)/s∞is obtained as ψ =N (s)N (0) −1Q 1X(s)Q1X(0)−1− 1 s ∞, (23)
where Q1X(s) = (X(s) + Di(s)Q1(s)). Thus the optimal ψo is the smallest ψ over
Q1X(s) = (X(s) + Di(s)Q1(s))for Q1∈ H∞. Define
Q1X(s) =:
No−1(s)
and " is the relative degree of No(s). Then, we have an invertible relation between the free
parameters Q1Xand QXinH∞. Note that the problem (23) is exactly in the form (13) except that Q1X(s)is restricted to have Q1X(pi) = X(pi) = 1/N (pi), whereas in (13) there is no
such restriction on the free parameter Qi∈ H∞. In summary, we have the following
ψ(QX) :=s−1( Ni(s) (1 + εs) QX(s) QX(0)− 1)∞ (24) ψo= inf{ ψ(QX) : QX ∈ H∞ and QX(pi) = (1 + εpi) Ni(pi) , i = 1 . . . , n}. (25) As in Section 3.1, we will be restricting ourselves to QX ∈ H∞such that QX(0) = 1,
because ψ(KQX) = ψ(QX) for any non-zero K. Thus, in the unstable plants case the
problem is modified to finding
ψo = inf QX 44 44s−1 Ni(s) (1 + εs)QX(s)− 1 44 44 ∞ subject to QX∈ H∞and QX(pi) = (1 + εpi) Ni(pi) , i = 1 . . . , n.
For a given γ > ψo, the set of all QX∈ H∞satisfying
44 44s−1 Ni(s) (1 + εs)QX(s)− 1 44 44 ∞ ≤ γ (26) can be characterized as Qγ= QX(s) = F1(s) + F2(s)U (s) F3(s) + F4(s)U (s) : U ∈ H∞, U∞≤ 1 , (27) where F1, . . . , F4are computed explicitly from the problem data, see e.g. [6]. Therefore, the problem at hand can be transformed to finding the smallest γ for which there exists U ∈ H∞,
U∞≤ 1 such that
F1(pi) + F2(pi)U (pi) F3(pi) + F4(pi)U (pi) = (1 + εpi) Ni(pi) =: αi, (28)
for i = 1 . . . , n. This leads to a set of interpolation conditions on U
U (pi) =
αiF3(pi)− F1(pi)
F2(pi)− αiF4(pi)
=: βi (29)
for i = 1 . . . , n. For each fixed γ we can find βi using for example [6]. Now we need to
check whether there exists U ∈ H∞ withU∞ ≤ 1 such that U(pi) = βi. This is a
Nevanlinna-Pick interpolation problem and it can be solved from the given problem data
{(p1. . . , pn), (β1, . . . , βn)}, see e.g. [6, 11, 18].
In summary, for unstable plants the problem is solved in two steps:
1. Given γ > ψo, solve the suboptimal version (26) of the problem (13) studied in
Sec-tion 3.1; characterize all suboptimal soluSec-tions in the form (27), i.e. find F1, F2, F3, F4. 2. Given p1, . . . , pn, determine β1, . . . , βnfrom the first step. Use this data to check if the
Nevanlinna-Pick interpolation problem has a feasible solution. If yes decrease γ, if no increase γ, and repeat Steps 1 and 2; using a bisection in this iteration find the optimal
γo. For γ = γo+ , where > 0, the Nevanlinna-Pick problem gives a solution U , which
Example. Let h > 0, and consider P (s) = e −hs s + 2 s + 1 + 2(s− 1)e−2s s + 3− 5e−0.5s . This plant is unstable with single pole p1= 0.6367inC+. Therefore, its factorization can be done as (20) where Di(s) = s− 0.6367 s + 0.6367 Ni(s) = e−hs (s + 1) + 2(s− 1)e−2s (1− s)e−2s− 2(s + 1) No(s) = (1− s)e−2s− 2(s + 1) (s + 2)(s + 3− 5e−0.5s) Di(s).
For h = 3 we have ψo = 8.6744. This gives α1 = Ni(p1)−1 = −14.945. Applying
the procedure described above we find F1, . . . , F4for each fixed γ > ψo, and compute β1
defined by (29). Since we have single interpolation condition, the solution of Nevanlinna-Pick problem is rather trivial: it is solvable if and only if|β1| ≤ 1, and as a solution we can take
U (s) = β1. By using a bi-section search we find that smallest γ > ψoleading to|β1| ≤ 1 is
γo= 13.4485, which leads to β1=−1, see Figure 3. Thus if we choose U(s) = −1, we get
QX(s) =
F1(s)− F2(s)
F3(s)− F4(s),
where F1, . . . , F4are computed from the solution of the suboptimal one-blockH∞problem
with γ = 13.45 > γo. 8 10 12 14 16 18 20 0 1 2 3 4 5 6 7 8 9 γ |β | |β| versus γ gives: γo=13.4485 Fig. 3. |β| versus γ.
4
Conclusions
A sufficient condition is derived for C = C1Cpi, cascade connection of a PI controller Cpi,
and an initial stabilizing controller C1, to stabilize a given plant P . Design of C1 for the largest allowable range of the integral action gain interval is investigated for stable plants, including systems with internal and input-output delays. We used parametrization of all sta-bilizing controller to characterize C1. Then we have seen that the problem at hand reduces to a weighted sensitivity minimization for a stable plant whose inner part is infinite dimensional and the weight is an integrator. When we consider a PD-like ˜Qi(s)in the parametrization of
C1, the problem becomes finding optimal kdin (18), which is solved in [14].
For unstable plants the problem of finding the largest allowable range of the integral action gain is solved in two steps. First the a suboptimal one-block problem is solved and in the second step a Nevanlinna-Pick interpolation problem is solved.
We should also point out that the result stated in Section 2 is a sufficient condition. There-fore the largest allowable integral action gain found in Section 3 is within the set of allowable gains characterized by this sufficient condition, which may be conservative. It would be inter-esting to investigate the level of conservatism in this approach. We leave this open problem to a future study.
APPENDIX
Proof of Theorem 1: a) Let P = ˜Y−1X˜ be a left-coprime-factorization (LCF) of P and let N1D−11 be a right-coprime-factorization (RCF) of C1. Since C1 stabilizes P , M1 :=
˜
Y D1+ ˜XN1 is unimodular inHr×r∞ . With C1∈ S(P ), we have Q1:= C1(I + P C1)−1∈
Hr×r
∞ and T1 := P C1(I + P C1)−1 = P Q1 ∈ Hr∞×r. Now Cy = I− Cxstabilizes T1
if and only if Cy(I + T1Cy)−1 ∈ H∞r×r, which implies (I + T1Cy)−1 ∈ Hr∞×r. Define
Dc:= (I + T Cy)−1D1, Nc= N1+ Q1CyDc; then Nc, Dc∈ Hr∞×r. Write C = C1Cx
as C = C1+ C1Cy= NcD−1c . Then ˜Y Dc+ ˜XNc= ˜Y Dc+ ˜X[N1+ Q1CyDc] = ˜Y Dc+
˜
XN1+ ˜Y P Q1CyDc= ˜Y (I +T1Cy)Dc+ ˜XN1= ˜Y (I +T1Cy)(I +T Cy)−1D1+ ˜XN1=
M1is unimodular and hence, C = C1Cx∈ S(P ).
b) Let P = XY−1be an RCF and C1= ˜D−11 N˜1 be an LCF. Then C1 ∈ S(P ) if and only if ˜M1 := ˜D1Y + ˜N1Xis unimodular; hence, det ˜M1(0)= 0. Since P, C1∈ S(P ) do not have transmission-zeros at s = 0, det X(0) = 0 and det ˜N1(0) = 0. Since det T1(0) = det X(0) ˜M1(0) ˜N1(0)= 0, we conclude that T1∈ Hr∞×rdoes not have transmission-zeros at s = 0. It follows from [9], Proposition 2, that the PID-controller ˆCpid in (4) stabilizes
T1∈ Hr∞×r. Therefore, by (1), C1Cx∈ S(P ), where Cx= I + ˆCpid.
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